# A Little Spring…

This one I wanted to do for a while. I like very much the math that can be derived from the Quantum Harmonic Oscillator, but there’s also a lot of stuff that I personally don’t like about the approach that is taken in some quantum physics books. I prefer a rather algebraic analysis of this system since I think that it is pretty much cleaner and easier to justify; then again I’m not pretending to be absolutely formal, just enough for it not to look like some kind of magic trick in the middle of the thing.

Consider the quantum system described by the Hamiltonian corresponding to an harmonic oscillator

$\displaystyle H=\frac{1}{2m}p^2 + \frac{1}{2}m\omega^2 x^2$

We can avoid writing all those constants by defining two new linear operators $P=\frac{p}{\sqrt{2m}}$ and $X=\sqrt{\frac{m}{2}}\omega x$. This allows us to write a much simpler Hamiltonian

$\displaystyle H=P^2+X^2 \qquad (*)$

We know that $[x,p] = i\hbar$, thus the commutation relationship for this new operators is given by $[X,P] = \frac{i}{2}\hbar\omega$. Notice that this new operators are also hermitian, as the momentum and position operators.

If we where to take the equation $(*)$ as classical variables, we could factorize the sum using complex conjugates as $a^2+b^2=(a+ib)(a-ib)$, but this relies on the commutativity of $a$ and $b$. Anyway we can go and try to do it. Consider the operator $z=X+iP$ and its adjoint $z^{\dagger}=X-iP$ (this last one follows from the hermiticity of $X$ and $P$). Since those are new operators, it may be useful to know its commutation relationship

$\displaystyle [z,z^{\dagger}]=[X+iP,X-iP]$

$\displaystyle [z,z^{\dagger}]=-i[X,P]+i[P,X]$

$\displaystyle [z,z^{\dagger}]=\frac{\hbar \omega}{2}+\frac{\hbar \omega}{2}=\hbar \omega$

Now, it is conventional to use a normalized version of the $z$ and $z^\dagger$ operators to get an unitary commutator. Define the annihilation and creation operators respectively as follows:

$\displaystyle a=\frac{X+iP}{\sqrt{\hbar\omega}} \qquad a^\dagger=\frac{X-iP}{\sqrt{\hbar\omega}}$

The reason for this names while be studied soon enough. For now the important thing is that its commutators is much cleaner: $[a,a^{\dagger}]=1$. Now we can proceed with this idea of multiplying “complex conjugates”:

$\displaystyle a^\dagger a=\frac{1}{\hbar \omega}(X-iP)(X+iP)$

$\displaystyle a^\dagger a=\frac{1}{\hbar \omega}(X^2+P^2+iXP-iPX)$

$\displaystyle a^\dagger a=\frac{1}{\hbar \omega}(H+i[X,P])= \frac{1}{\hbar \omega}\left(H-\frac{1}{2}\hbar\omega\right)$

Solving for the Hamiltonian we get the desired form:

$\displaystyle H=\hbar \omega \left(a^\dagger a +\frac{1}{2}\right) (\odot)$

Alternatively, we could write $\displaystyle H=\hbar \omega \left(a a^\dagger-\frac{1}{2}\right)$, but there is something to notice before advancing any further.

Lemma 1. Let $A$ be a linear operator acting on and one of its eigenpairs $\{\lambda, \left|\varphi\right> \}$ (i.e. $A \left|\varphi\right>=\lambda\left|\varphi\right>$ ). For a given constant $c$ $\{\lambda +c , \left|\varphi\right> \}$ is an eigenpair of the linear operator $(A + c)$.

Proof. Follows trivially from adding $c\left|\varphi\right>$ to both sides of the equation $A \left|\varphi\right>=\lambda\left|\varphi\right> \quad \square$.

From linear algebra we know that $a^\dagger a$ is a Positive-Definite Operator; thus, every eigenvalue of this operator are positive. This implies, using the Lemma 1, the following observation:

Observation 1. The eigenvalues of the Harmonic Oscillator Hamiltonian operator have a lower bound of $\frac{\hbar \omega}{2}$.

Now that we know a little more about this Hamiltonian and that we have already written it down as a function of the creation and annihilation operators, it is possible to calculate the commutator of the Hamiltonian with the annihilation operator

$[H,a]=\hbar \omega[a^\dagger a, a]$

$\displaystyle [H,a]=\hbar \omega(a^\dagger [a,a]+[a^\dagger,a]a)$

$[H,a]=-\hbar \omega a \qquad (\triangle)$

Take now an eigenpair $\{\lambda, \left|\psi\right>\}$ of the Hamiltonian. We study the action of the Hamiltonian on the vector space defined by $\left|a\psi\right> \equiv a\left|\psi\right>$ using the equation $(\triangle)$:

$H\left|a\psi\right> = Ha\left|\psi\right>$

$H\left|a\psi\right>=(aH -\hbar \omega a)\left|\psi\right>$

$H\left|a\psi\right>=(a\lambda - \hbar \omega a)\left|\psi\right>$

$H\left|a\psi\right>=(\lambda - \hbar \omega)a\left|\psi\right>= (\lambda - \hbar \omega)\left|a\psi\right> \qquad (\circledast)$

This implies that the annihilation operator acting on an eigenvector, generates a new eigenstate with an lower eigenvalue by an amount of $\hbar \omega$. In an analogous manner we can prove that $[H,a^\dagger]=\hbar \omega a^\dagger$ and consequently that $H\left|a^\dagger\psi\right>=(\lambda + \hbar \omega)\left|a^\dagger\psi\right>$.

From the previous we see that from given an eigenpair of this Hamiltonian, we can get a discrete non-degenerated set of eigenpairs. Even thought we can do this for every eigenpair, it should be clear that the sets generated by two different do not necessarily coincide one with the other, but all of these sequences should have a minimum element, because of the Observation 1. We now consider now one of those minima, its eigenstate will be denote as $\left| 0 \right>$. Applying the annihilation operator to these state should generate a null vector due to the minimality of $\left| 0 \right>$. Writing these operator explicitly allows us to know the general form of the state:

$a\left|0\right>=0$

$\left(\frac{X+iP}{\sqrt{\hbar\omega}}\right)\left|0\right>=0$

$\left(\sqrt{\frac{m}{2}}\omega x+\frac{\hbar}{\sqrt{2m}}\partial_x\right)\left|0\right>=0$

$\frac{\mathrm{d}\left|0\right>}{\mathrm{d} x}= -\frac{m\omega x}{\hbar} \left|0\right>$

$\int \frac{\mathrm{d}\left|0\right>}{\left|0\right>}= -\int\frac{m\omega x}{\hbar}\mathrm{d}x$

$\left|0\right>=Ce^{-\frac{m\omega}{2\hbar}x^2}$

Now, remember that the square magnitude of this vector state is to represent a probability density and should therefore be normalized

$\left<0|0\right>=|C|^2\int_{-\infty}^{\infty}e^{-\frac{m\omega}{\hbar}x^2}\mathrm{d}x$

$\left<0|0\right>=|C|^2\sqrt{\frac{\pi \hbar}{m\omega}}$

$|C|=\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}$

With this we finally get the desired state

$\left|0\right>=\left(\frac{m\omega}{\pi \hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^2}$

Moreover, notice that this is an unique solution and therefore we have a definite lowest energy level; since every sequence of eigenpairs generated by an initial one should get to a minimum through applications of the annihilation operators, we see that every sequence should get to this particular state $\left|0\right>$ and therefore there’s a unique discrete sequence of eigenstates generated by the annihilation and creation operators. From now on we shall refer to $\left|0\right>$  as the ground state of the harmonic oscillator.

Using the representation given by $(\odot)$ we can get the corresponding energy for the ground state

$H\left|0\right>=\hbar \omega \left( a^\dagger a +\frac{1}{2} \right)\left|0\right>$

$H\left|0\right>=\hbar \omega \left( 0 +\frac{1}{2} \right)\left|0\right>$

$H\left|0\right>=\frac{\hbar \omega}{2}\left|0\right>$

Applying the creation operator $n$ times in the ground state vector, we reach a vector on the ray corresponding to the $n^{th}$ state of the system. Each application increases the corresponding eigenvalue leaving us with the set of eigenpairs described by $\left\{\hbar \omega \left(n+\frac{1}{2} \right),\left|n\right>\right\}$.

Something important to notice here is that $a$ and  $a^\dagger$ do not generate vector states, since we never actually checked them to be normalized, but instead we get vectors on the ray corresponding to this state. We now see the actual effect of these operators on vector states:

$a^\dagger\left|n\right>=\alpha\left|n+1\right> \qquad \Rightarrow \left

$\left=|\alpha|^2 \left = |\alpha|^2$

$\left=|\alpha|^2$

$|\alpha|^2= \left$

$|\alpha|= \sqrt{n+1}$

This steps (and some analogous ones for the annihilation operator) leaves us the following rules:

$a^\dagger\left|n\right>=\sqrt{n+1}\left|n+1\right> \qquad a\left|n\right>=\sqrt{n}\left|n-1\right>$

And also a general way to generate eigenstates of the Hamiltonian operator:

$\left|n\right>=\frac{(a^\dagger)^n}{\sqrt{n!}} \left|0\right>$

There is one more thing to be done here: The Matrix Representation. Well, lets not jump that far yet, first we need to talk about inner products in this Hilbert space. Consider the differential equation associated to the eigenvalue problem of the system (Time-independent Schrödinger equation):

$-\frac{\hbar ^2}{2m}\frac{\mathrm{d}^2\psi }{\mathrm{d} x^2} + \frac{1}{2} m\omega^2 x^2 \psi = E \psi$

This equation is in the Sturm-Liouville (I know there’s something dirty going on here, mainly having to do with boundary conditions and stuff which I will ignore, because YOLO). This tells us that the normalized solutions for this equations are orthonormal:

$\left=\delta_{nm}$

Furthermore, the are a complete set, meaning that every state can be written as a linear combination of those eigenstates.

Note: in case this is not clear, here we are using the following inner product definition: $\left<\varphi|\psi\right>=\int_{-\infty}^{\infty}\mathrm{d}x \bar{\varphi}\psi$

From definition we can write the position and momentum operators as a linear combination of the creation and annihilation operators:

$x=\sqrt{\frac{\hbar}{2m\omega}}(a^\dagger+a) \qquad p=i\sqrt{\frac{m\hbar \omega}{2}}(a^\dagger-a) \qquad (\circledast)$

This allows us to get a much more nice way to obtain the matrix representation of such operators

$x \rightarrow x_{mn}=\left=\sqrt{\frac{\hbar}{2m\omega}}\left$

$x \rightarrow x_{mn}=\sqrt{\frac{\hbar}{2m\omega}}\left(\left+\left\right)$

$x \rightarrow x_{mn}=\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n+1}\left+\sqrt{n}\left\right)$

$x \rightarrow x_{mn}=\sqrt{\frac{\hbar}{2m\omega}}\left(\delta_{m,n+1}\sqrt{n+1}+\delta_{m,n-1}\sqrt{n}\right)$

Analogously, for the momentum operator we get

$p \rightarrow p_{mn}=i\sqrt{\frac{m\hbar \omega}{2}}\left(\delta_{m,n+1}\sqrt{n+1}-\delta_{m,n-1}\sqrt{n}\right)$

This form of writing this may be a little awkward to interpret, but here I leave you the matrix form:

$x \rightarrow \sqrt{\frac{\hbar}{2m\omega}}\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots\\ \sqrt{1} & 0 & \sqrt{2} & 0 & \cdots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \cdots \\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \end{pmatrix}$

$p \rightarrow i\sqrt{\frac{m\hbar \omega}{2}}\begin{pmatrix} 0 & -\sqrt{1} & 0 & 0 & \cdots\\ \sqrt{1} & 0 & -\sqrt{2} & 0 & \cdots \\ 0 & \sqrt{2} & 0 & -\sqrt{3} & \cdots \\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \end{pmatrix}$

And the creation and annihilation look something like this:

$a \rightarrow \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \cdots\\ 0 & 0 & \sqrt{2} & 0 & \cdots \\ 0 & 0 & 0 & \sqrt{3} & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \end{pmatrix}$

$a^\dagger \rightarrow \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots\\ \sqrt{1} & 0 & 0 & 0 & \cdots \\ 0 & \sqrt{2} & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{3} & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \end{pmatrix}$

Done! There might be other stuff that can be done in this system, like analyzing the uncertainty of the position and momentum of a given eigenstate (Notice from the matrix representation that the expectation value for both observables is zero, but this is not the case for its standard deviations), nevertheless this does not creates anything new to the things we have been doing, but it rather uses what we have already been doing. Anyway, it might be a nice exercise to calculate $\sigma_x^2$ and $\sigma_p^2$ from $(\circledast)$.

For the sake of completeness, I leave here the expression for a general state, considering the time evolution of the system, I invite the reader to verify that this is in fact the case:

$\left|\psi\right>= e^{\frac{i\omega t}{2}}\sum_{k=0}^{\infty}\alpha_n e^{in\omega t}\left|n\right>, \qquad \alpha_n\in \mathbb{C}$

And this is it! I invite you to play a little more with this system by yourself to see any additional feature, or check the “classical” approach to it (solving the differential equation by conventional methods) in case you haven’t see it before so you can convince yourself that this method is much cleaner. For now I will leave it here now that I have taken this out of my chest.

Have a nice day.